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The maximum possible value of $xy^2z^3$ subjected to condition $x,y,z \geq 0$ and $x+y+z=3$ is

  1. $1$
  2. $\frac{9}{8}$
  3. $\frac{9}{4}$
  4. $\frac{27}{16}$
asked in Calculus by Boss (17.9k points)
recategorized by | 332 views

1 Answer

+17 votes
Best answer

Given ,

$x + y + z = 3$

$\Rightarrow  x +\left(\frac{y}{2}\right) +\left(\frac{y}{2}\right) +\left(\frac{z}{3}\right) +\left(\frac{z}{3}\right) +\left(\frac{z}{3}\right) = 3.$

Now using A.M. G.M inequality , we have : 

$\dfrac{\left[x +\left(\frac{y}{2}\right) +\left(\frac{y}{2}\right) +\left(\frac{z}{3}\right) +\left(\frac{z}{3}\right) +\left(\frac{z}{3}\right)\right]}{6}$

 $\geq \large\left[x .\left(\frac{y}{2}\right) .\left(\frac{y}{2}\right) .\left(\frac{z}{3}\right) .\left(\frac{z}{3}\right) .\left(\frac{z}{3}\right)\right]^{\frac{1}{6}}$

$\Rightarrow  \large\left(x .\left(\frac{y}{2}\right) .\left(\frac{y}{2}\right) .\left(\frac{z}{3}\right) .\left(\frac{z}{3}\right) .\left(\frac{z}{3}\right)\right)^{\frac{1}{6}}\leq \frac{1}{2}$

$\Rightarrow  \large\left(x .\left(\frac{y}{2}\right) .\left(\frac{y}{2}\right) .\left(\frac{z}{3}\right) .\left(\frac{z}{3}\right) .\left(\frac{z}{3}\right)\right)\leq \frac{1}{64}$

$\Rightarrow x . y^{2} . z^{3}  \leq \frac {108}{64}  =\frac {27}{16}.$

Hence  D) is the correct answer

answered by Veteran (100k points)
edited by
+1
We can also solve this question by lagrange method of undetermined multiplier
0
which rule applied here from x+y+z=3 to x+y/2+y/2+z/3+z/3+z/3=3.i cant undersatand this step plz explain it
0
y=y/2+y/2=2y/2=y

z=z/3+z/3+z/3=3z/3=z
0

@Tesla!

how this AM and GM inequality is applied??

am not getting it!!pls explain!!

0

@Gate Fever

its straight forward AM of x,y,z = $\frac{x+y+z}{3}$

and GM of X,Y,Z is $\sqrt[3]{x+y+z}$

for purpose of this question we are making some changes in AM and GM and using the inequality 

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